/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 A camera lens \((n=1.50)\) is co... [FREE SOLUTION] | 91影视

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Chapter 25: Problem 19

A camera lens \((n=1.50)\) is coated with a thin film of magnesium fluoride \((n=1.38)\) of thickness \(90.0 \mathrm{nm}\) What wavelength in the visible spectrum is most strongly transmitted through the film?

Short Answer

Expert verified
Answer: The most strongly transmitted wavelength through the magnesium fluoride film is approximately 686 nm.

Step by step solution

01

Identify relevant formulas and concepts

In a thin film of material, light waves will interact with each other, causing interference. For constructive interference to occur, the additional path length travelled by light waves through the film must be an integer multiple of the wavelength. The general formula for constructive interference in a thin film is given by: \(2nt=m位'\), where \(n\) is the refractive index of the film material, \(t\) is the thickness of the film, \(m\) is an integer multiple, and \(位'\) is the wavelength of light in the film. The relationship between the wavelength of light in the film \(位'\) and the wavelength of light in air (or vacuum) \(位\) is given by \(位'=\frac{位}{n}\).
02

Calculate the path length

We are given the refractive index of magnesium fluoride (\(n = 1.38\)) and the thickness of the film (\(t = 90.0\,\text{nm}\)). Using these values, we can calculate the path length, \(2nt\), by simply multiplying them together, and then multiplying by 2. Path length, \(2nt = 2 脳 1.38 脳 90.0\,\text{nm} = 248.4\,\text{nm}\).
03

Determine the constructive interference condition

Recall that the constructive interference condition is given by \(2nt=m位'\). We already have the path length \(2nt=248.4\,\text{nm}\), so we will use \(位'=\frac{位}{n}\) to rewrite the condition in terms of the wavelength of light in air. We get \(248.4\,\text{nm} = m\frac{位}{1.38}\).
04

Choose the visible wavelength range

The visible spectrum of light consists of wavelengths from about \(380\,\text{nm}\) to \(750\,\text{nm}\). We need to find the wavelength in this range that is most strongly transmitted through the film.
05

Determine the integer multiple, m

In order to find the most strongly transmitted wavelength, we need to find the integer multiple, \(m\), that satisfies the constructive interference condition (\(248.4\,\text{nm} = m\frac{位}{1.38}\)) for a wavelength within the visible spectrum. We can solve for \(m\) and then for \(位\): \(m=\frac{248.4\,\text{nm}脳1.38}{位}\). In this case, we will use an iterative method, testing values of \(m\) until we find the one that will give us a wavelength in the visible range.
06

Calculate the most strongly transmitted wavelength

By using different integer values for \(m\) and calculating the corresponding wavelengths \(位\), we find that when \(m=2\), the wavelength is within the visible range: \(位 = \frac{248.4\,\text{nm}脳1.38}{2} 鈮� 686\,\text{nm}\). This wavelength is within the visible range (between \(380\,\text{nm}\) and \(750\,\text{nm}\)) and close to the red end of the spectrum. So the most strongly transmitted wavelength through the magnesium fluoride film is approximately \(686\,\text{nm}\).

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Most popular questions from this chapter

A steep cliff west of Lydia's home reflects a \(1020-\mathrm{kHz}\) radio signal from a station that is \(74 \mathrm{km}\) due east of her home. If there is destructive interference, what is the minimum distance of the cliff from her home? Assume there is a \(180^{\circ}\) phase shift when the wave reflects from the cliff.
The radio telescope at Arecibo, Puerto Rico, has a reflecting spherical bowl of \(305 \mathrm{m}(1000 \mathrm{ft})\) diameter. Radio signals can be received and emitted at various frequencies with appropriate antennae at the focal point of the reflecting bowl. At a frequency of \(300 \mathrm{MHz}\), what is the angle between two stars that can barely be resolved? (Tutorial:radio telescope).
A Michelson interferometer is set up using white light. The arms are adjusted so that a bright white spot appears on the screen (constructive interference for all wavelengths). A slab of glass \((n=1.46)\) is inserted into one of the arms. To return to the white spot, the mirror in the other arm is moved $6.73 \mathrm{cm} .$ (a) Is the mirror moved in or out? Explain. (b) What is the thickness of the slab of glass?
Light incident on a pair of slits produces an interference pattern on a screen \(2.50 \mathrm{m}\) from the slits. If the slit separation is $0.0150 \mathrm{cm}$ and the distance between adjacent bright fringes in the pattern is \(0.760 \mathrm{cm},\) what is the wavelength of the light? [Hint: Is the small-angle approximation justified?]
A grating spectrometer is used to resolve wavelengths \(660.0 \mathrm{nm}\) and \(661.4 \mathrm{nm}\) in second order. (a) How many slits per centimeter must the grating have to produce both wavelengths in second order? (The answer is either a maximum or a minimum number of slits per centimeter.) (b) The minimum number of slits required to resolve two closely spaced lines is $N=\lambda /(m \Delta \lambda),\( where \)\lambda$ is the average of the two wavelengths, \(\Delta \lambda\) is the difference between the two wavelengths, and \(m\) is the order. What minimum number of slits must this grating have to resolve the lines in second order?
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